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Mathematical Induction EECS 203: Discrete Mathematics Lecture 11 Spring 2016 1

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Climbing the Ladder We want to show that ∀ n≥1 P(n) is true. –Think of the positive integers as a ladder. – 1, 2, 3, 4, 5, 6,... You can reach the bottom of the ladder: – P(1) From each ladder step, you can reach the next. – P(1) → P(2), P(2) → P(3),... – ∀ k≥1 P(k) → P(k+1) Then, by mathematical induction: – ∀ n≥1 P(n) 2

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Mathematical Induction How do we prove a universal statement about the positive integers: ∀ n≥1 P(n) Mathematical induction is an inference rule –Base case: P(1) –Inductive step: ∀ k≥1 P(k) → P(k+1) –Conclusion: ∀ n≥1 P(n) The inductive step requires proving an implication. 3

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Simple Math Induction Example Prove, for all n ∈ N, Proof: We use induction. Let P(n) be: Base case: P(0), because 0 = 0. 4

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Simple Math Induction Example Inductive step: Assume that P(k) is true, where k is an arbitrary natural number. 5

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Simple Math Induction Example Inductive step: Assume that P(k) is true, where k is an arbitrary natural number. The first equality follows from P(k), and the second by simplification. Thus, P(k+1) is true. By induction, P(n) is true for all natural numbers n. QED –This shows how to write a clear inductive proof. 6

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Question Did you understand the proof by induction? –(A) Yes, that’s really clever! –(B) How can you start with P(0) instead of P(1)? –(C) When do you use P(k) and when P(n)? –(D) Isn’t it circular to assume P(k), when you are trying to prove P(n)? –(E) I just don’t get it. 7

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Divisibility Example (1) Let’s try to prove ∀ n ∈ N 3 | n 3 −n. –First, we’ll work through the mathematics. –Then we’ll structure a clear proof. Base case: P(0) is easy: 3 | (0 3 −0) –Everything divides zero, so it’s true. 8

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Divisibility Example (2) Inductive step: ∀ k ∈ N P(k) → P(k+1) –Assume P(k): 3 | (k 3 −k) (for arbitrary k) –Can we prove P(k+1): 3 | ((k+1) 3 – (k+1)) ? Try multiplying it out: We know that 3 | (k 3 −k) and 3 | (3k 2 +3k). Conclude that 3 | ((k+1) 3 – (k+1)), so P(k) → P(k+1). 9

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A Clear Divisibility Proof Theorem: ∀ n ∈ N 3 | (n 3 −n) Proof: Use induction on P(n): 3 | (n 3 −n). Base case: P(0) is true because 3 | (0 3 −0). Inductive step: Assume P(k), for arbitrary natural number k. Then: The final statement is P(k+1), so we have proved that P(k) → P(k+1) for all natural numbers k. By mathematical induction, P(n) is true for all n ∈ N. QED 10

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Notice: we had to go back and start by assuming P(k) and then showing it implied P(k+1). –That first step isn’t at all clear unless you’ve done the math first! 11

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Theorem: Proof: Base case: Inductive step: 12

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A False “Proof” 13

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False “Proof” (2) Inductive hypothesis P(k): In any set of k horses, all have the same color. Now consider a set of k+1 horses – h 1, h 2,... h k, h k+1 By P(k), the first k horses are all the same color – h 1, h 2,... h k, Likewise, the last k horses are also the same color – h 2,... h k, h k+1 Therefore, all k+1 horses must have the same color Thus, ∀ k ∈ N P(k) → P(k+1) So, ∀ n ∈ N P(n). What’s wrong? 14

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Question What’s wrong with that proof by induction? –(A) Mathematical induction doesn’t apply to horses. –(B) The Base Case is incorrect. –(C) The Inductive Step is incorrect. –(D) The Base Case and Inductive Step are correct, but you’ve put them together wrong. –(E) It’s completely correct: all horses are the same color! 15

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The Problem with the Horses Inductive step: ∀ k ∈ N P(k) → P(k+1) –Now consider a set of k+1 horses h 1, h 2,... h k, h k+1 –By P(k), the first k horses are all the same color h 1, h 2,... h k, –Likewise, the last k horses are also the same color h 2,... h k, h k+1 –Therefore, all k+1 horses must have the same color 16 A set of one horse with the same color Another set of one horse with the same color

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The Problem with the Horses 17

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The Problem with the Horses Inductive step: ∀ k ∈ N P(k) → P(k+1) –Now consider a set of k+1 horses h 1, h 2,... h k, h k+1 –By P(k), the first k horses are all the same color h 1, h 2,... h k, –Likewise, the last k horses are also the same color h 2,... h k, h k+1 –Therefore, all k+1 horses must have the same color 18

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The Problem with the Horses Inductive step: ∀ k ∈ N P(k) → P(k+1) –Now consider a set of k+1 horses h 1, h 2,... h k, h k+1 –By P(k), the first k horses are all the same color h 1, h 2,... h k, –Likewise, the last k horses are also the same color h 2,... h k, h k+1 –Therefore, all k+1 horses must have the same color 19 A group of two

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The Problem with the Horses Inductive step: ∀ k ∈ N P(k) → P(k+1) –Now consider a set of k+1 horses h 1, h 2,... h k, h k+1 –By P(k), the first k horses are all the same color h 1, h 2,... h k, –Likewise, the last k horses are also the same color h 2,... h k, h k+1 –Therefore, all k+1 horses must have the same color 20 Another group of two

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The Problem with the Horses Inductive step: ∀ k ∈ N P(k) → P(k+1) –Now consider a set of k+1 horses h 1, h 2,... h k, h k+1 –By P(k), the first k horses are all the same color h 1, h 2,... h k, –Likewise, the last k horses are also the same color h 2,... h k, h k+1 –Therefore, all k+1 horses must have the same color 21 But base step wasn’t true!

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Tiling a Checkerboard (1) For n≥1, consider a 2 n ×2 n checkerboard. Can we cover it with 3-square L-shaped tiles? No. Remove any one of the squares. Can we tile it now? Prove by induction: Let P(n) be the proposition: –A 2 n ×2 n checkerboard, minus any one of the squares, can be tiled with these L-shaped tiles. 22

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Tiling a Checkerboard (2) Consider a 2 k+1 ×2 k+1 checkerboard, minus any one square. –Divide the checkerboard into four 2 k ×2 k quadrants. The missing square is in one quadrant. –From the other three quadrants, remove the square closest to the center of the checkerboard. 23

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Tiling a Checkerboard (3) P(k) says that each quadrant (minus one square) can be tiled. One tile covers the three central squares. Thus, P(k) → P(k+1), for arbitrary k≥1. By mathematical induction, ∀ n≥1 P(n). 24

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The Harmonic Series (1) The Harmonic Numbers are the partial sums of the Harmonic Series: We want to prove that: This implies an important property of the Harmonic Series: 25

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The Harmonic Series (2) Theorem: Proof: By induction, with P(n) being Base case: P(0) is true because Inductive step: The inductive hypothesis P(k) is From that assumption, we want to prove P(k+1). 26

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The Harmonic Series (3) The inductive hypothesis P(k) is This demonstrates P(k+1). 27

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The Harmonic Series (4) This proves that P(k) P(k+1) for arbitrary k. Therefore, by mathematical induction, P(n) is true for all n N. QED. –Theorem: Let j = 2 n be some integer. Then n = log j. –Therefore, the Harmonic series diverges, because –But it diverges very slowly. 28

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Prove that 6 divides n 3 − n whenever n is a nonnegative integer. 29

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